1.7 KiB
一、第一问
\because 圆中同弧所对的圆周角相等
\therefore \angle AED=\angle ABD ①
连接AD, $\because \angle ADB$是直径$AB$所对的圆周角，\therefore \angle ADB=90^{\circ}
又 $\because D$是$BC$中点，所以\triangle ADB \cong \triangle ADC
\therefore \angle ABD = \angle ACB ②
联立 ① ②, \therefore \angle AED=\angle C
二、第二问
\because \angle AED=55^{\circ}
\therefore \angle ABC=\angle C=55^{\circ}
\therefore \angle BAC=180^{\circ}-55^{\circ}-55^{\circ}=70^{\circ}
根据 圆的内接四边形对角互补,$\therefore \angle BDF=110^{\circ}$
三、第三问
看到线段乘积，考虑找到相似三角形
\large \frac{EG}{?}=\frac{?}{ED}
考虑\triangle EAG \sim \triangle EAD
如何证明呢？
有一个公共角\angle AED
还需要再找一个角：
因为$E$是 \overset{{\frown}}{AB} 的中点，
\therefore \angle ADE=\angle BAE
\therefore \triangle EAG \sim \triangle EAD
\large \frac{EG}{AE}=\frac{AE}{ED}
\Rightarrow EG*ED=AE^2
$AE$长度如何求解呢？
\because \angle AFD+\angle ABD=180^{\circ}
\because \angle AFD+\angle CFD=180^{\circ}
\therefore \angle ABD=\angle CFD
因为有第一问的结论，所以$\triangle DCF$是等腰三角形，CD=DF=BD=4
\because cos\angle ABD=\frac{2}{3}
\large \therefore \frac{BD}{AB}=\frac{2}{3}
\therefore AB=6
$\because \triangle ABE$是等腰直角三角形
\therefore AE=3\sqrt{2}
\therefore EG*ED=(3\sqrt{2})^2=18